# [LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历创建二叉树

2021年09月15日 阅读数：1

Given preorder and inorder traversal of a tree, construct the binary tree.html

Note:
You may assume that duplicates do not exist in the tree.java

For example, givennode

```preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]```

Return the following binary tree:python

```    3
/ \
9  20
/  \
15   7```

Java:ui

```public TreeNode buildTree(int[] preorder, int[] inorder) {
return helper(0, 0, inorder.length - 1, preorder, inorder);
}

public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
if (preStart > preorder.length - 1 || inStart > inEnd) {
return null;
}
TreeNode root = new TreeNode(preorder[preStart]);
int inIndex = 0; // Index of current root in inorder
for (int i = inStart; i <= inEnd; i++) {
if (inorder[i] == root.val) {
inIndex = i;
}
}
root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);
return root;
}
```

Python: Recursiveurl

```def buildTree(self, preorder, inorder):
if inorder:
ind = inorder.index(preorder.pop(0))
root = TreeNode(inorder[ind])
root.left = self.buildTree(preorder, inorder[0:ind])
root.right = self.buildTree(preorder, inorder[ind+1:])
return root
```

Python:htm

```class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution:
# @param preorder, a list of integers
# @param inorder, a list of integers
# @return a tree node
def buildTree(self, preorder, inorder):
lookup = {}
for i, num in enumerate(inorder):
lookup[num] = i
return self.buildTreeRecu(lookup, preorder, inorder, 0, 0, len(inorder))

def buildTreeRecu(self, lookup, preorder, inorder, pre_start, in_start, in_end):
if in_start == in_end:
return None
node = TreeNode(preorder[pre_start])
i = lookup[preorder[pre_start]]
node.left = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1, in_start, i)
node.right = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1 + i - in_start, i + 1, in_end)
return node

if __name__ ==  "__main__":
preorder = [1, 2, 3]
inorder = [2, 1, 3]
result = Solution().buildTree(preorder, inorder)
print(result.val)
print(result.left.val)
print(result.right.val)
```

C++:blog

```// Time:  O(n)
// Space: O(n)

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
unordered_map<int, size_t> in_entry_idx_map;
for (size_t i = 0; i < inorder.size(); ++i) {
in_entry_idx_map.emplace(inorder[i], i);
}
return ReconstructPreInOrdersHelper(preorder, 0, preorder.size(), inorder, 0, inorder.size(),
in_entry_idx_map);
}

// Reconstructs the binary tree from pre[pre_s : pre_e - 1] and
// in[in_s : in_e - 1].
TreeNode *ReconstructPreInOrdersHelper(const vector<int>& preorder, size_t pre_s, size_t pre_e,
const vector<int>& inorder, size_t in_s, size_t in_e,
const unordered_map<int, size_t>& in_entry_idx_map) {
if (pre_s == pre_e || in_s == in_e) {
return nullptr;
}

auto idx = in_entry_idx_map.at(preorder[pre_s]);
auto left_tree_size = idx - in_s;

auto node = new TreeNode(preorder[pre_s]);
node->left = ReconstructPreInOrdersHelper(preorder, pre_s + 1, pre_s + 1 + left_tree_size,
inorder, in_s, idx, in_entry_idx_map);
node->right = ReconstructPreInOrdersHelper(preorder, pre_s + 1 + left_tree_size, pre_e,
inorder, idx + 1, in_e, in_entry_idx_map);
return node;
}
};
```

[LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历创建二叉树